杭电2054-A == B ?

Myth丶恋晨 2022-08-19 02:16 140阅读 0赞

# A == B ? #

**Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)**

Total Submission(s): 83216 Accepted Submission(s): 13071

Problem Description

Give you two numbers A and B, if A is equal to B, you should print "YES", or print "NO".

Input

each test case contains two numbers A and B.

Output

for each case, if A is equal to B, you should print "YES", or print "NO".

Sample Input

1 2
    2 2
    3 3
    4 3

Sample Output

NO
    YES
    YES
    NO

卡了很长时间的题了，要考虑很多种情况，感谢大牛数据！

测试数据
    -00001==-1
    -0==0
    10.000==10
    00010==10
    10.==10 
    */

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    char s1[100100],s2[100100];
    int main()
    {
        int len1,len2,m,n,a,b,i,j,flag1,flag2,t,f1,f2;
        while(scanf("%s%s",s1+1,s2+1)!=EOF)
        {
            len1=strlen(s1+1);
            len2=strlen(s2+1);
            flag1=flag2=f1=f2=0;
            if(s1[0]=='-')
            flag1=1;
            if(s2[0]=='-')
            flag2=1;
            s1[0]='\0';
            s2[0]='\0';
            
            for(i=1;i<=len1;i++)
            {
                if(s1[i]=='.')
                f1=1;
            }
            for(i=1;i<=len2;i++)
            {
                if(s2[i]=='.')
                f2=1;
            }
    
                for(i=1;i<=len1;i++)
                {
                    if(s1[i+1]=='.'||(s1[i]>'0'&&s1[i]<='9'))
                    break;
                }
                j=len1;
                if(f1)
                for(;s1[j]=='0'&&j>=0;j--)
                ;
                if(s1[j]=='.')
                j--;
                t=0;
                for(a=i;a<=j;a++)
                s1[t++]=s1[a];
                s1[t]='\0';
        
                for(i=1;i<=len2;i++)
                {
                    if(s2[i+1]=='.'||(s2[i]>'0'&&s2[i]<='9'))
                    break;
                }
                j=len2;
                if(f2)
                for(;s2[j]=='0'&&j>=0;j--)
                ;
                if(s2[j]=='.')
                j--;
                t=0;
                for(a=i;a<=j;a++)
                s2[t++]=s2[a];
                s2[t]='\0';
                
            
            if((flag1+flag2)%2==0&&strcmp(s1,s2)==0)
            printf("YES\n");
            else
            printf("NO\n");
        }
        return 0;
    }